Thermodynamics Part 3

Given, initial load, $M_1=185\ \text{kg}$ Initial pressure, $p_1=76\ \text{cm}$ of Hg Initial temperature, $T_1=27^{\circ}\text{C}=(273+27)\text{K}=300\text{K}$ Final pressure at height, $p_2=45\ \text{cm}$ of Hg Final temperature, $T_2=-7^{\circ}\text{C}=[273+(-7)]\text{K}=266\text{K}$ Volume of balloon is constant, $V_1=V_2=V$ We know that, for air, $p=\rho RT$ Now, for initial condition, we can write $\rho_1=\rho_1 RT_1$ ...(i) Similarly, for final condition, we can write $P_2=\rho_2 RT_2$ ...(ii) Dividing Eq. (i) by Eq. (ii), we get $\frac{p_1}{p_2}=\frac{\rho_1 T_1}{\rho_2 T_2}$ Substituting the given values in above expression, we get $\frac{76\ \text{cm of Hg}}{45\ \text{cm of Hg}}=\frac{\rho_1}{\rho_2}\left(\frac{300\ \text{K}}{266\ \text{K}}\right)$ $\Rightarrow\ \frac{\rho_1}{\rho_2}=\frac{76\times266}{45\times300}$ We know that, density of balloon can be written as $\frac{\rho M}{V}$ As, volume is constant ρ ∝ M We can write, $\frac{\rho_1}{\rho_2}=\frac{M_1}{M_2}=\frac{76\times266}{45\times300}$ $\Rightarrow \frac{M_1}{M_2}=\frac{20216}{13500}$ Substituting the value of $M_1$ in above expression, we get $M_2=185\times\frac{13500}{20216}=123.54\ \text{kg}$ Thus, the load carried by balloon will be 123.54 kg.